📚 IB Physics: Cambridge Formula Derivations | IB 物理：剑桥公式推导
Mastering the derivations behind key equations is essential for success in IB Physics. Understanding where formulas come from not only deepens your conceptual grasp but also equips you to tackle unfamiliar problems with confidence. This article walks through the most important derivations in the Cambridge IB Physics syllabus, from kinematics to electricity, providing clear step‐by‐step reasoning with full bilingual support.
掌握关键方程背后的推导对于 IB 物理的成功至关重要。理解公式的来源不仅能加深你的概念理解，还能让你有信心解决不熟悉的问题。本文将从运动学到电磁学，逐步讲解剑桥 IB 物理大纲中最重要的推导，提供清晰的双语解释。
1. Deriving the SUVAT Equations | 运动学基本公式推导
The SUVAT equations relate displacement (s), initial velocity (u), final velocity (v), constant acceleration (a) and time (t). They originate from the definition of acceleration and the area under a velocity–time graph.
SUVAT 方程将位移 (s)、初速度 (u)、末速度 (v)、恒定加速度 (a) 和时间 (t) 联系起来。它们源自加速度的定义以及速度-时间图下的面积。
Acceleration is the rate of change of velocity: a = (v – u) / t. Rearranging gives the first equation: v = u + a t. Since acceleration is constant, the average velocity is (u + v)/2, and displacement is average velocity times time: s = ((u + v)/2) t. Substituting v = u + a t into this expression yields s = u t + ½ a t². Eliminating t from v = u + a t and s = ((u + v)/2) t gives v² = u² + 2 a s.
加速度是速度的变化率：a = (v – u) / t。整理后得到第一个方程：v = u + a t。由于加速度恒定，平均速度为 (u + v)/2，位移等于平均速度乘以时间：s = ((u + v)/2) t。将 v = u + a t 代入该式得到 s = u t + ½ a t²。从 v = u + a t 和 s = ((u + v)/2) t 中消去 t，可得 v² = u² + 2 a s。
2. Projectile Motion: Range and Maximum Height | 抛体运动：射程与最大高度
Projectile motion is analysed by resolving the initial velocity into horizontal and vertical components. Under negligible air resistance, the horizontal motion is uniform while the vertical motion has constant downward acceleration g.
抛体运动通过将初速度分解为水平和竖直分量来分析。在空气阻力可忽略的情况下，水平方向做匀速运动，竖直方向受恒定向下的加速度 g。
Let a projectile be launched with speed u at angle θ to the horizontal. Horizontal component: uₓ = u cos θ; vertical component: u_y = u sin θ. Time to reach maximum height is found when vertical velocity becomes zero: 0 = u sin θ – g t, so t_peak = (u sin θ) / g. The total time of flight is twice this: T = (2 u sin θ) / g. Range R = horizontal velocity × T = (u cos θ) × (2 u sin θ / g) = (u² sin 2θ) / g. Maximum height H is obtained from v_y² = u_y² – 2 g y, setting v_y = 0: H = (u² sin² θ) / (2 g).
设物体以速度 u、与水平成角 θ 抛出。水平分量：uₓ = u cos θ；竖直分量：u_y = u sin θ。到达最大高度时竖直速度为零：0 = u sin θ – g t，因此 t_peak = (u sin θ) / g。总飞行时间是其两倍：T = (2 u sin θ) / g。射程 R = 水平速度 × T = (u cos θ) × (2 u sin θ / g) = (u² sin 2θ) / g。最大高度 H 由 v_y² = u_y² – 2 g y，令 v_y = 0 得到：H = (u² sin² θ) / (2 g)。
3. Newton’s Second Law and Momentum | 牛顿第二定律与动量
Newton’s second law is often stated as F = m a, but its most fundamental form uses momentum. Momentum p is defined as p = m v. The net force acting on an object equals the rate of change of its momentum.
牛顿第二定律通常表述为 F = m a，但其最基本的形式使用动量。动量 p 定义为 p = m v。作用在物体上的合外力等于其动量的变化率。
Starting from F = Δp / Δt, if mass is constant, Δp = m Δv, so F = m (Δv / Δt) = m a. When mass changes (e.g. a rocket), the full derivative F = dp/dt = d(mv)/dt = m (dv/dt) + v (dm/dt) must be used. This derivation also leads to the impulse–momentum theorem: impulse J = F Δt = Δp.
从 F = Δp / Δt 开始，如果质量恒定，Δp = m Δv，因此 F = m (Δv / Δt) = m a。当质量变化时（如火箭），必须使用全微分 F = dp/dt = d(mv)/dt = m (dv/dt) + v (dm/dt)。这一推导也引出了冲量-动量定理：冲量 J = F Δt = Δp。
4. Derivation of Kinetic Energy and Work–Energy Theorem | 动能和工作-能量定理的推导
The work done by a net force on an object causes a change in its kinetic energy. This can be derived from Newton’s second law and the definition of work.
净力对物体做的功导致其动能的变化。这可以从牛顿第二定律和功的定义推导出来。
Consider a constant net force F acting over a displacement s. Work done: W = F s. Using F = m a and the kinematic equation v² = u² + 2 a s, we replace a s: a s = (v² – u²) / 2. Then W = m × (v² – u²)/2 = ½ m v² – ½ m u². Hence W_net = ΔK, where kinetic energy K = ½ m v². This is the work–energy theorem. It also holds for variable forces when using integration: W = ∫ F·dx = ∫ m v dv = ½ m v² – ½ m u².
考虑一个恒定的净力 F 作用在位移 s 上。做功：W = F s。利用 F = m a 和运动学方程 v² = u² + 2 a s，替换 a s：a s = (v² – u²) / 2。于是 W = m × (v² – u²)/2 = ½ m v² – ½ m u²。因此 W_net = ΔK，其中动能 K = ½ m v²。这就是功-能定理。对于变力，利用积分同样成立：W = ∫ F·dx = ∫ m v dv = ½ m v² – ½ m u²。
5. Centripetal Acceleration | 向心加速度推导
An object moving in a circle of radius r at constant speed v experiences an acceleration directed towards the centre. This is derived from the geometry of circular motion.
一个物体以恒定速率 v 在半径为 r 的圆周上运动时，会受到指向圆心的加速度。这是从圆周运动的几何性质推导出来的。
In a short time Δt, the object moves from position A to B, subtending an angle Δθ. The change in velocity Δv points towards the centre, and its magnitude is |Δv| = v Δθ (since the velocity vector rotates by Δθ). The distance traveled is v Δt = r Δθ, so Δθ = (v Δt) / r. Therefore, acceleration magnitude a = |Δv| / Δt = v (v Δt / r) / Δt = v² / r. Using v = ω r, it can also be written as a = ω² r.
在很短的时间 Δt 内，物体从位置 A 运动到 B，转过的角度为 Δθ。速度的变化量 Δv 指向圆心，其大小为 |Δv| = v Δθ（因为速度矢量旋转了 Δθ）。经过的距离为 v Δt = r Δθ，因此 Δθ = (v Δt) / r。于是加速度大小 a = |Δv| / Δt = v (v Δt / r) / Δt = v² / r。利用 v = ω r，也可写成 a = ω² r。
6. Simple Harmonic Motion: Period of a Spring–Mass System | 简谐运动：弹簧-振子周期
A spring–mass system executes simple harmonic motion (SHM) when Hooke’s law applies. The period T can be derived from the second-order differential equation for SHM.
当胡克定律适用时，弹簧-振子系统做简谐运动（SHM）。周期 T 可以从 SHM 的二阶微分方程推导出来。
Hooke’s law: F = –k x. By Newton’s second law, m a = –k x, so a = –(k/m) x. Since acceleration is the second derivative of displacement, d²x/dt² = –(k/m) x. The general solution is x = A cos(ω t + φ) with ω = √(k/m). Angular frequency ω = 2π / T, hence T = 2π √(m/k). For a simple pendulum, the restoring force leads to ω = √(g/L), giving T = 2π √(L/g).
胡克定律：F = –k x。由牛顿第二定律，m a = –k x，因此 a = –(k/m) x。由于加速度是位移的二阶导数，d²x/dt² = –(k/m) x。通解为 x = A cos(ω t + φ)，其中 ω = √(k/m)。角频率 ω = 2π / T，因此 T = 2π √(m/k)。对于单摆，回复力导出 ω = √(g/L)，得到 T = 2π √(L/g)。
7. Kinetic Theory: Pressure of an Ideal Gas | 分子动理论：理想气体压强
The pressure exerted by an ideal gas is a macroscopic result of countless molecular collisions with the container walls. By considering momentum change, we derive p = (1/3) ρ ⟨c²⟩ or pV = N m ⟨c²⟩ / 3.
理想气体产生的压强是无数分子与容器壁碰撞的宏观结果。通过考虑动量变化，可以推导出 p = (1/3) ρ ⟨c²⟩ 或 pV = N m ⟨c²⟩ / 3。
Imagine a cubic container of side L, containing N molecules each of mass m. Focus on one molecule moving towards a wall with velocity component vₓ. Momentum change per collision is 2 m vₓ. The time between successive collisions with that wall is 2L / vₓ. Force on the wall = rate of momentum change = (2 m vₓ) / (2L / vₓ) = m vₓ² / L. Summing over all molecules, total force F = (m / L) Σ vₓ². Pressure p = F / L² = (m / L³) Σ vₓ² = (m / V) N ⟨vₓ²⟩. Because the motion is random, ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = (1/3) ⟨c²⟩, where c is the speed. Hence p = (1/3) (N m / V) ⟨c²⟩, or pV = (1/3) N m ⟨c²⟩ = (2/3) N ⟨KE⟩.
设想边长为 L 的立方体容器，内有 N 个质量为 m 的分子。考虑一个分子以速度分量 vₓ 朝一壁运动。每次碰撞的动量变化为 2 m vₓ。与该壁连续两次碰撞的时间间隔为 2L / vₓ。壁上的力 = 动量变化率 = (2 m vₓ) / (2L / vₓ) = m vₓ² / L。对所有分子求和，总力 F = (m / L) Σ vₓ²。压强 p = F / L² = (m / L³) Σ vₓ² = (m / V) N ⟨vₓ²⟩。由于运动是随机的，⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = (1/3) ⟨c²⟩，其中 c 为速率。因此 p = (1/3) (N m / V) ⟨c²⟩ 或 pV = (1/3) N m ⟨c²⟩ = (2/3) N ⟨KE⟩。
8. Coulomb’s Law and Electric Field Strength | 库仑定律与电场强度
Coulomb’s law describes the force between two point charges. From this experimentally verified law, we can derive the expression for electric field strength and the potential around a point charge.
库仑定律描述了两个点电荷之间的力。从这个已被实验验证的定律出发，我们可以推导出点电荷周围的电场强度和电势表达式。
Coulomb’s law: F = k q₁ q₂ / r², where k = 1/(4π ε₀). The electric field strength E is defined as the force per unit positive charge: E = F / q. For a point charge Q, the force on a test charge q is F = k Q q / r², so E = k Q / r². The electric potential V at distance r is the work done per unit charge to bring a small charge from infinity to that point: V = –∫ᵣ∞ E dr = –∫ᵣ∞ (k Q / r²) dr = k Q / r.
库仑定律：F = k q₁ q₂ / r²，其中 k = 1/(4π ε₀)。电场强度 E 定义为每单位正电荷所受的力：E = F / q。对于点电荷 Q，施加在检验电荷 q 上的力为 F = k Q q / r²，因此 E = k Q / r²。距离 r 处的电势 V 是将一小电荷从无穷远移至该点每单位电荷所做的功：V = –∫ᵣ∞ E dr = –∫ᵣ∞ (k Q / r²) dr = k Q / r。
9. Resistors in Series and Parallel | 电阻的串联与并联
By applying conservation of charge and energy to circuits, we can derive the equivalent resistance for resistors connected in series and in parallel.
通过对电路应用电荷守恒和能量守恒，我们可以推导出串联和并联电阻的等效电阻。
Series: The same current I flows through each resistor. The total potential difference across the combination is the sum of individual p.d.s: V = V₁ + V₂ + V₃ … = I R₁ + I R₂ + I R₃ … = I (R₁ + R₂ + R₃ …). Hence the equivalent resistance R_series = R₁ + R₂ + R₃ …. Parallel: The p.d. V is the same across each branch. The total current I = I₁ + I₂ + I₃ … = V/R₁ + V/R₂ + V/R₃ … = V (1/R₁ + 1/R₂ + 1/R₃ …). Thus 1/R_parallel = 1/R₁ + 1/R₂ + 1/R₃ ….
串联： 相同的电流 I 流过每个电阻。组合两端的总电势差是各个电势差之和：V = V₁ + V₂ + V₃ … = I R₁ + I R₂ + I R₃ … = I (R₁ + R₂ + R₃ …)。因此等效电阻 R_series = R₁ + R₂ + R₃ …。并联： 电势差 V 在每个支路上相同。总电流 I = I₁ + I₂ + I₃ … = V/R₁ + V/R₂ + V/R₃ … = V (1/R₁ + 1/R₂ + 1/R₃ …)。因此 1/R_parallel = 1/R₁ + 1/R₂ + 1/R₃ …。
10. Faraday’s Law of Electromagnetic Induction | 法拉第电磁感应定律
Faraday’s law states that the induced emf in a coil equals the negative rate of change of magnetic flux linkage. We can derive this fundamental law by considering Lorentz forces on moving charges in a conductor.
法拉第定律指出，线圈中感应的电动势等于磁通量链变化率的负值。我们可以通过考虑作用于导体中运动电荷的洛伦兹力来推导这一基本定律。
Consider a straight conductor of length L moving with velocity v perpendicular to a uniform magnetic field B. Free electrons in the conductor experience a magnetic force F = q v B (Lorentz force). This force separates charges, building up an electric field until equilibrium: q E = q v B, so E = v B. The induced emf across the conductor is ε = E L = B L v. In terms of flux, the area swept per time is L v, so the flux change dΦ = B dA = B L v dt, giving ε = dΦ/dt. For a coil of N turns, flux linkage NΦ gives ε = –N dΦ/dt (Lenz’s law sign).
考虑一根长度为 L 的直导体，以速度 v 垂直于均匀磁场 B 运动。导体中的自由电子受到磁力 F = q v B（洛伦兹力）。该力使电荷分离，建立起电场直至平衡：q E = q v B，因此 E = v B。导体两端的感应电动势为 ε = E L = B L v。就磁通量而言，单位时间扫过的面积是 L v，因此磁通量变化 dΦ = B dA = B L v dt，从而 ε = dΦ/dt。对于 N 匝线圈，磁链 NΦ 给出 ε = –N dΦ/dt（楞次定律中的负号）。
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